Asked by Anonymous
Find the point on the curve y = 1/x^2 ,( x > 0 ), where the curvature is maximum.
Answers
Answered by
oobleck
k = |y"|/(1+y'^2)^(3/2) = (6/x^4)/(1 + 4/x^6)^(3/2)
so, maximum k is where
24(5-x^6)/(x^6+4)^(5/2) = 0
That is, at x = 5^(1/6)
That would make the point (5^(1/6),5^(-1/3))
so, maximum k is where
24(5-x^6)/(x^6+4)^(5/2) = 0
That is, at x = 5^(1/6)
That would make the point (5^(1/6),5^(-1/3))
There are no AI answers yet. The ability to request AI answers is coming soon!
Submit Your Answer
We prioritize human answers over AI answers.
If you are human, and you can answer this question, please submit your answer.